of 1 44
An Abstract Theory of Reality
By
Ian Beardsley
Copyright © 2024 by Ian Beardsley
of 2 44
Contents
Preface…………………………………………………………………4
The Spacetime Operators……………………………………….5
Solving The Atom………………………………………………….11
Solving The Earth Orbit…………………………………………13
The Solar Formulation………………………………………….17
Equating The Lunar And Solar
Formulations Yield Our 1 Second Base Unit……………20
Characteristic Time of the Electron……………………….24
Intermediate Mass And Velocity…………………………….26
Characteristic Time Of Electron Gives The
Moon’s Diameter………………………………………………….30
Biological Life Part of a Universal Natural Process
In The Universe……………………………………………………31
The Data For Verifying The Equations……………………….43
of 3 44
The main job of finding a theory of everything (TOE) is to make the pieces fit together. That is
the primary task, which can be achieved in an abstract sense, leaving the results open to
interpretation. We often see theories are reformulated, but many of the equations are the same,
and this may be because the equations are from measurables, but since mathematics is self-
referential, abstract, and an approximation to reality because of its assumptions which result in
conundrums, we are left with reality ending in paradox. It may be reality is an infinite regress, so
making the pieces fit together suffices, and the interpretations can be left to the state of
knowledge at the time, which is always changing.
Here I formulate spacetime operators that solve the atom and Earth/Moon/Sun System in
common. It may be that they solve more than that, but that is where I have applied them here. It
may be many more operators are needed for solutions of everything. The operators are in
kilograms/meters/seconds because space and time (meters and seconds) define matter
(kilograms) and everything, or most everything, seems to be a relationship between mass, space,
and time.
I find the base unit of Nature is 1 second in duration, which is interesting, if not strange, because
the second came from a base 60 sexagesimal system of counting handed to us by the ancient
Sumerians who started civilization some 12,000 years ago, long before knowledge of the atom
and wave equations. I find I can solve the Earth/Moon/Sun system with the Schrödinger wave
equation used for proton/electron systems comprising the atom, and that the quantization is
done in terms of the Moon, which seems to be some sort of a Natural Yardstick.
In order to solve the electron in these terms I have to introduce the concept of intermediate
mass and intermediate velocity. The result is we go from the particles that make atoms, to the
atoms, which are the elements, then to the planets and the Sun, for a journey from the
microscale to the macroscale where everything unfolds as one.
Finally, in this second edition there is a paper at the end that shows our base unit of a second for
atomic and planetary systems also explains hydrocarbons, the skeletons of biological life.
of 4 44
Preface
You can speak of the structure of the solar system even though it changes with time. This is
important to understand when I refer to sizes of the Moon and the planets, and their orbital
distances.
The whole object of developing a theory for the way planetary systems form is that they meet the
following criterion: They predict the Titius-Bode rule for the distribution of the planets; the
distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The
distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from
the parent star, the outward pressure gradient from the stellar production of radiation, and the
outward inertial forces as a cloud collapses into a flat disc around the central star. These forces
separate the flat disc into rings, agglomerations of material, each ring from which a different
planet forms at its central distance from the star (it has a thickness). In a theory of planetary
formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of
planets.
of 5 44
The Spacetime Operators
In my paper The Solar System As A Solution To The Wave Equation (Beardsley, 2024) I
showed that I could solve the solar system, like we do with the atom, with the quantum
mechanical Schrödinger wave equation and that the result is done in terms of the Moon and a
base unit of one second. The solar system is as such connected to the proton of the atom. As a
result I found that
1)
Where is the radius of a proton, is the mass of a proton, is the fine structure constant,
is Planck’s constant, is the constant of gravitation, and is the speed of light. We see this by
considering
2)
=
=
Where . The units
3)
describe the Nature of space and time. We say 2E-12 of it are the degree to which space acts on
the size and mass of a proton to give it a characteristic time of one second:
4)
We see that same unit of time is in the orbital energies (kinetic energies) of the Earth and the
Moon in terms of the Earth’s spin, its day, or period of rotation which is
1secon d =
1
6α
2
⋅
r
p
m
p
4πh
Gc
r
p
m
p
α
h
G
c
1
6α
2
4πh
Gc
(3128.167)
4π (6.62607E − 34)
(6.67408E − 11)(299,792,458)
2.018E − 12
kg ⋅ s
m
≈ 2E − 12
kg ⋅ s
m
α = 1/137
kg ⋅ s
m
r
p
m
p
=
0.833E − 15m
1.67262E − 27kg
= 4.9802E11
m
kg
≈ 5E11
m
kg
1
6α
2
4πh
Gc
⋅
r
p
m
p
=
(
5E11
kg ⋅ s
m
)(
2E − 12
m
kg
)
= 1secon d
(Ear th Da y) = (24hrs)(60min)(60sec) = 86400secon ds
of 6 44
That is the solar day, the rotation of the Earth with respect to the Sun. It makes little difference
whether we use it or, the sidereal day, the rotation of the Earth with respect to the stars. We
have
5)
Where we have used the orbital velocity of the Moon at aphelion, and that of the Earth at
perihelion, though it makes little difference on how you do it because the orbits are close to
circular. We get
Where KE means kinetic energy and subscripts m and e are moon and earth respectively.
Why does the basis of our counting system, 1 second, happen to be the basis unit of time for our
solution of the solar system and the atom. I don’t know, but I can say how we ended-up with that
unit for the basis of our counting.
The interesting thing is that we have the duration of a second is what it is not because we were
attempting to find some base unit of time for Nature, but rather we have the duration of a
second from ancient times that came to us over many years of dividing the rotation of the Earth
into hours, minutes, and seconds in such a way that we ended up with it. However we may have
had some sort of a thought going into it that was six-fold in Nature that lead it to being natural if
we consider:
6)
And that thought may be was in a six-fold pattern that the motions of the Earth/Moon/Sun
system had in it approximately which we inadvertently characterized in the calendar:
7)
Which can be written
8)
Where there are 360 degrees in a circle and as such the Earth approximately moves through one
degree per day around the Sun. We achieved this because the ancients of Sumeria and Babylonia
from which the ancient Greeks received their counting system used base 60, sexagesimal
because 60 is evenly divisible by so much, that is by
9) 1,2,3,4,5,6, 10, 12, 15, 20, 30,…
1secon d =
K E
m
K E
e
(Ear th Da y)
K E
m
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
1secon d =
1
365.25
⋅
1
24
⋅
1
60
⋅
1
60
= 0.0000000316881year
(
360d a ys
year
)(
24h ours
d a y
)(
60mi n
h our
)(
60sec
min
)
= 31104000secon ds /year
12
2
(
60d a ys
year
)(
1h our
d a y
)(
60mi n
h our
)(
60sec
min
)
of 7 44
The Sumerians divided the day into 12 hours, evenly divisible into 60, and so from sunrise to
sunrise was 24 hours, and from sunset to sunset was 24 hours. They multiplied 60 by 6, to get
360 degrees in a circle. Since the Earth does one orbit around the Sun in 365 days, which is a
year, the Earth goes through about 1 degree a day. The Babylonians got that from them, and
further divided the hour into 60 minutes, and the minute into 60 seconds, though hours were in
their sun dials, they only used minutes and seconds for astronomy. The West, the Europeans,
got their sun dials from the Babylonians but did not incorporate minutes and seconds into their
mechanical clocks until the 17th century, but it all came from the Sumerians, who settled down
from hunting with stone spearpoint and started agriculture, writing, mathematics, and
civilization in general. Sun dials were used in Egypt as well and had 12 divisions throughout the
day as well for hours. A sun dial simply tells time by the angle of a shadow cast by the Sun as the
sun rises and sets, its position measuring time.
We have one second in equations 1, and 5:
Equation 5 can be written
10)
Since orbital velocities are given by
We have equation 10 is
11)
It can be made more accurate using aphelions and perihelions appropriately. It is an amazing
fact that the Moon as seen from the Earth near perfectly eclipses the Sun. To say that
mathematically is to write
12)
1secon d =
1
6α
2
⋅
r
p
m
p
4πh
Gc
1secon d =
K E
m
K E
e
(Ear th Da y)
M
m
v
2
m
M
e
v
2
e
(Ear th Da y) = 1secon d
v
2
m
= G
M
e
r
m
v
2
e
= G
M
⊙
r
e
M
m
M
⊙
⋅
r
e
r
m
(Ear th Da y) = 1secon d
7.347673E 22kg
1.989E 30kg
⋅
1.496E11m
3.84E8m
(86,400s) = 1.243secon ds
r
e
r
m
=
R
⊙
R
m
of 8 44
That is, the earth orbital radius to that of the Moon is equal to the radius of the Sun to the radius
of the Moon. Putting that into equation 11 yields:
13)
Let us look at equation 4 and write it:
14)
And write 13 as
15)
So that we see ( means corresponds to)
M
m
M
⊙
⋅
R
⊙
R
m
(Ear th Da y) = 1secon d
7.347675E 22kg
1.989E 30kg
⋅
6.96E8m
1,737,400m
(86,400s) = 1.2786secon d s
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
(
M
m
R
m
(Ear th Da y)
)
⋅
R
⊙
M
⊙
= 1secon d
≜
of 9 44
And
So, we have
16)
17)
Thus the spacetime operator for the Earth-Moon system is on the order of 1.5E33 times greater
than that for the atom, and the radius of the proton to the mass of the proton is on the order
order of 1.5E33 times greater than that of the radius of the Sun to its mass. Thus the two
systems, that of the atom and that of the Earth and Moon nearly balance with C1/C2 about equal
to one.
It is worth noting here, while the Moon near perfectly eclipses the Sun, and that this was seen by
prehistoric man that hunted with stone spearpoint, the ratio of the the radius of the Sun to the
orbital radius of the Moon is about 9/5=1.8, which the ratio of the molar mass of gold to the
molar mass of silver, that is it is the mass of a gold atom to a silver atom. This is interesting
because gold and silver have been the primary metals used since ancient times for ceremonial
jewelry, gold yellow like the Sun, silver, silver like the Moon. We have
is the radius of the Sun, is the orbital radius of the Moon, Au, is gold and Ag is silver.
So, not only did the Sumerians of Mesopotamia, who created writing, mathematics, and
agriculture, divide the day into 12 units called hours, so did the Ancient Egyptians. They both
measured hours with sun dials, simply a stick that casts a shadow from the Sun that moves as
(
1
6α
2
4πh
Gc
)
= 2.018E − 12
kg ⋅ s
m
≜
(
M
m
R
m
(Ear th Da y)
)
= 3.654E 21
kg ⋅ s
m
r
p
m
p
= 4.98E11
m
kg
≜
R
⊙
M
⊙
= 3.5E − 22
m
kg
C
1
(
1
6α
2
4πh
Gc
)
=
(
M
m
R
m
(Ear th Da y)
)
r
p
m
p
= C
2
R
⊙
M
⊙
C
1
=
3.654E 21
2.018E − 12
= 1.8107E 33
C
2
=
4.98E11
3.5E − 22
= 1.423E 33
C
1
C
2
= 1.27245
R
⊙
r
m
=
Au
Ag
R
⊙
r
m
of 10 44
the Sun moves, rising in the East and setting in the West. As the shadow moves it passes
through twelve lines each line representing the hour of a day. We see from equation 8
that gives the seconds per year, that the 1 hour day exists naturally in a base 60 counting. The
Sumerians also used base 12, because it divides evenly into 60 and is an abundant number, full
of divisors, 1, 2, 3, 4, 6 all divide evenly into it and add up to 16, which greater than 12. It makes
computing very easy.
We have spacetime operators operating on radius to mass of a proton and operating on radius to
mass of the the Sun, equations 14 and 15:
The operator for the proton is in units of
But we can also make a spacetime operator in which acts not on but on to
give a characteristic time of one second as well. It is
18)
It acts on
19)
We have
12
2
(
60d a ys
year
)(
1h our
d a y
)(
60mi n
h our
)(
60sec
min
)
= secon d s /year
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
(
M
m
R
m
(Ear th Da y)
)
⋅
R
⊙
M
⊙
= 1secon d
(
1
6α
2
4πh
Gc
)
=
kg ⋅ s
m
s
m ⋅ kg
m /kg
kg ⋅ m
2
3
⋅
π r
p
α
4
G m
3
p
=
s
m ⋅ kg
1
3
⋅
h
p
c
= kg ⋅ m
2
3
⋅
π r
p
α
4
G m
3
p
=
2
3
⋅
(352275361)π (0.833E − 15m)
(6.67408E − 11)(1.67262E − 27kg)
3
= 1.402818E42
s
m ⋅ kg
1
3
⋅
h
c
=
1
3
⋅
6.62607E − 34
299,792,458
= 7.3674E − 43kg ⋅ m
of 11 44
We have
So we have equations 20, and 14:
20)
14)
We have that
And
Solving The Atom
We see the spacetime operators solve the atom by giving us the radius of a proton. We set
equation 20 equal to equation 14
20)
14).
These two yield
(
1.402818E42
s
m ⋅ kg
)
(
7.3674E − 43kg ⋅ m
)
= 1.0335secon ds = 1.0secon ds
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
p
c
= 1secon d
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
(
1
6α
2
4πh
Gc
)
= 2.018E − 12
kg ⋅ s
m
≈ 2E − 12
kg ⋅ s
m
r
p
m
p
= 4.9802E11
m
kg
≈ 5E11
m
kg
2
3
⋅
π r
p
α
4
G m
3
p
= 1.402818E42
s
m ⋅ kg
1
3
⋅
h
c
= 7.3674E − 43kg ⋅ m
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
of 12 44
21).
We can arrive at this radius of a proton another way, energy is given by Planck’s constant and
frequency
We have
We take the rest energy of the mass of a proton :
The frequency of a proton is
This gives
The radius of a proton is then
This is close to the CODATA value for the proton radius around 2018 (0.842E-15m). The result
is
We find it may be that the radius of a proton is actually
22)
r
p
=
2
3
⋅
h
cm
p
E = h f
f = 1/s, h = J ⋅ s, h f = (J ⋅ s)(1/s) = J
E = J = Joules = energ y
m
p
E = m
p
c
2
f
p
=
m
p
c
2
h
m
p
c
2
h
⋅
r
p
c
=
2
3
≈ ϕ =
m
p
c
h
r
p
m
p
r
p
=
2
3
⋅
h
c
r
p
=
2
3
⋅
h
cm
p
r
p
=
2
3
⋅
6.62607E − 34
(299,792,458)(1.67262E − 27)
= 0.88094E − 15m
r
p
= ϕ ⋅
h
cm
p
= 0.816632E − 15m
of 13 44
Where is the golden ratio constant (0.618…). This is more along the lines of more recent
measurements. Thus we have solved the atom with our spacetime operators by producing the
radius of a proton.
Solving The Earth Orbit
Looking at equations 14 and 15
14)
15)
They seem to suggest that the Earth orbit might by quantized in terms of the Moon and a base
unit of one second.
We say there is a Planck constant for the solar system. We do it with a base unit of one second
because equations 14 and 15 suggest we should . We suggest it is such that it is given by the
standard Planck constant for the atom, , times some constant, , and the Kinetic energy of the
Earth.
23)
24)
Where
25).
Where equation 25 comes from equation 20
20).
We derive the value of our solar Planck constant
=
ϕ
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
(
M
m
R
m
(Ear th Da y)
)
⋅
R
⊙
M
⊙
= 1secon d
h
C
ℏ
⊙
= (hC )K E
e
hC = 1secon d
C =
1
3
⋅
1
α
2
c
2
3
⋅
π r
p
G m
3
p
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
1
3
⋅
18769
299792458
1
3
⋅
2π (0.833E − 15)
(6.67408E − 11)(1.67262E − 27)
3
of 14 44
=
=
=
Now we show that our Planck constant for the solar system gives the base unit of one second for
the quantization. We call our solar Planck constant and find the wavelength for the Moon
which is the ground state for the solar system:
26)
Then wavelength associated with the Moon divided by the speed of light should be 1 second if
our planetary system is quantized in terms of the Moon and a base unit of one second. We have
27)
And we see it is, so we have
28).
The solution for the orbit of the Earth around Sun with the Schrödinger wave equation can be
inferred from the solution for an electron around a proton in the a hydrogen atom with the
Schrödinger wave equation. The Schrödinger wave equation is, in spherical coordinates
1.55976565E 33
s
m
m
kg
3
⋅
s
2
kg
m
3
=
s
m
s
2
kg
2
m
2
=
s
m
⋅
s
kg ⋅ m
=
1
kg
⋅
s
2
m
2
1
C
= kg
m
2
s
2
=
1
2
mv
2
= en erg y
hC = (6.62607E − 34)(1.55976565E 33) = 1.03351secon d s ≈ 1.0secon d s
hC =
(
kg
m
s
2
⋅ m ⋅ s
)
(
1
kg
⋅
s
2
m
2
)
(
kg
m
2
s
)(
1
kg
⋅
s
2
m
2
)
= secon d s
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
ℏ
⊙
= (hC )K E
earth
= (1.03351s)(2.7396E 33J ) = 2.8314E 33J ⋅ s
ℏ
⊙
λ
moon
=
ℏ
2
⊙
GM
3
m
=
(2.8314E 33)
2
(6.67408E − 11)(7.34763E 22kg)
3
= 3.0281E8m
λ
moon
c
=
3.0281E8m
299,792,458m /s
= 1.010secon ds
λ
moon
c
= 1secon d
of 15 44
29)
Its solution for the atom is
30)
31)
is the energy for an electron orbiting protons and , is the orbital shell for an electron with
protons, the orbital number. I find the solution for the Earth around the Sun utilizes the
Moon around the Earth. This is different than with the atom because planets and moons are not
all the same size and mass like electrons and protons are, and they don’t jump from orbit to
orbit like electrons do. I find that for the Earth around the Sun
32)
33)
is the kinetic energy of the Earth, and is the planet’s orbit. is the radius of the Sun,
is the radius of the Moon’s orbit, is the mass of the Earth, is the mass of the Moon, is
the orbit number of the Earth which is 3 and is the Planck constant for the solar system.
Instead of having protons, we have the radius of the Sun normalized by the radius of
the Moon. It is has always been an amazing fact the the sizes of the Moon and the Sun are such
that given their orbital distances, the Moon as seen from the Earth perfectly eclipses the Sun.
This becomes part of the theory and we suggest it is a condition for sophisticated planets that
harbor life by writing it:
34)
That is the orbital radius of the planet (Earth) to the orbital radius of its moon (The Moon) is
about equal to the the radius of the star (The Sun) to the radius of its moon (The Moon). We say
the system is quantized by the Moon and the base unit of one second. It is a fact that the Moon
orbiting the Earth optimizes the conditions for life on Earth because it holds the Earth at its
inclination to its orbit around the Sun allowing for the seasons, preventing extreme hot and
extreme cold. Let us now see if our solar Planck constant works…
35).
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sinθ
∂
∂θ
(
sinθ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
E = −
Z
2
(k
e
e
2
)
2
m
e
2ℏ
2
n
2
r
n
=
n
2
ℏ
2
Z k
e
e
2
m
e
E
Z
r
n
Z
n
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
r
n
=
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
K E
e
r
n
R
⊙
r
m
M
e
M
m
n
ℏ
⊙
Z
R
⊙
/R
m
r
planet
r
moon
≈
R
star
R
moon
R
⊙
R
m
=
6.96E8m
1737400m
= 400.5986
of 16 44
=
=2.727E36J
The kinetic energy of the Earth is
The accuracy of our equation is:
Which is very good. Thus we have solved the Earth/Moon/Sun System with our spacetime
operators by using them to find a Planck constant for the solar system.
The Basis Set For The Solar System
We want the basis set of equations for the solar system. We have equation 26
26)
We have from equation 27
36).
We have equations 23, and 24
23)
24)
We have equations 20, 14, and 5
20)
14)
5)
K E
e
= (1.732)(400.5986)
(6.67408E − 11)
2
(5.972E 24kg)
2
(7.347673E 22kg)
3
2(2.8314E 33)
2
K E
e
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.727E 36J
2.7396E 33J
100 = 99.5 %
λ
moon
=
ℏ
2
⊙
GM
3
m
= 3.0281E8m
λ
moon
c
=
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1.0secon ds
ℏ
⊙
= (hC )K E
e
hC = 1secon d
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
1secon d =
K E
m
K E
e
(Ear th Da y)
of 17 44
From these it becomes clear that
37)
38)
39).
The Solar Formulation
Our solution of the wave equation for the planets gives the kinetic energy of the Earth from the
mass of the Moon orbiting the Earth, but you could formulate based on the Earth orbiting the
Sun. In our lunar formulation we had:
40)
We remember the Moon perfectly eclipses the Sun which is to say
41)
Thus equation 40 becomes
42)
The kinetic energy of the Earth is
43)
Putting this in equation 42 gives the mass of the Sun:
44)
We recognize that the orbital velocity of the Moon is
45)
So equation 44 becomes
ℏ
⊙
=
GM
3
m
c
K E
e
K E
e
=
GM
3
m
c
ℏ
⊙
ℏ
⊙
=
(
1
6α
2
⋅
r
p
m
p
4πh
Gc
)
K E
e
K E
e
= 3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
R
⊙
R
m
=
r
e
r
m
K E
e
= 3
r
e
r
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
e
=
1
2
⋅
GM
⊙
M
e
r
e
M
⊙
= 3r
2
e
⋅
GM
e
r
m
⋅
M
3
m
ℏ
2
⊙
v
2
m
=
GM
e
r
m
of 18 44
46)
This gives the mass of the Moon is
47)
Putting this in equation 40 yields
48)
We now multiply through by and we have
49)
Thus the Planck constant for the Sun, , in this the case the star is the Sun, is angular
momentum quantized as Bohr discovered for the atom by The angular
momentum we will call , the subscript for Planck. We have
We write for the solution of the Earth/Sun system:
50)
Let us compare this to that of an atom:
51)
We notice that in equation 50
; ; ; ;
is really . We can write 50 as
M
⊙
= 3r
2
e
⋅ v
2
m
⋅
M
3
m
ℏ
2
⊙
M
3
m
=
M
⊙
ℏ
2
⊙
3r
2
e
v
2
m
K E
e
=
R
⊙
R
m
⋅
G
2
M
2
e
M
⊙
2r
2
e
v
2
m
M
2
e
/M
2
e
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2r
2
e
v
2
m
M
2
e
ℏ
⊙
h /2,2h /2,3h /2,...
L
p
p
L
p
= r
e
v
m
M
e
= r
e
v
m
M
e
= (1.496E11m)(1022 m /s)(5.972E 24kg) = 9.13E 38kg ⋅
m
2
s
L
2
p
= r
2
e
v
2
m
M
2
e
= 7.4483E 77J ⋅ m
2
⋅ kg = 8.3367E 77kg
2
⋅
m
4
s
2
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
E = −
Z
2
n
2
⋅
k
2
e
e
4
m
e
2ℏ
2
Z
2
n
2
→
R
⊙
R
m
k
2
e
→ G
2
e
4
→ M
4
e
m
e
→ M
⊙
ℏ
2
→ L
2
p
L
p
ℏ
⊙
of 19 44
52)
We say. . That is
Let us see how accurate our equation is:
=
=
We have that the kinetic energy of the Earth is
Our equation has an accuracy of
Which is very good.
We call the same in our solar solution. But we now want our solution for the solar
formulation. is the mass of the Earth and is the mass of the Sun. We have our solution
might be!
53) !
Where is !
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
ℏ
⊙
= ℏ
⊙
/2π
ℏ
⊙
= 9.13E 38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39J ⋅ s
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
R
⊙
R
m
(6.67408E − 11)
2
(5.972E 24kg)
4
(1.9891E 30kg)
2(8.3367E 77kg
2
⋅
m
4
s
2
)
R
⊙
R
m
(6.759E 30J )
R
⊙
R
m
=
6.957E8m
1737400m
= 400.426
K E
e
= 2.70655E 33J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
2.70655E 33J
2.7396E 33J
= 98.79 %
L
p
ℏ
⊙
r
n
M
e
M
⊙
r
n
=
ℏ
2
⊙
GM
3
e
⋅
R
m
R
⊙
h
⊙
of 20 44
We have
54)
This has an accuracy of
!
Thus the solutions to the wave equation!
!
for the solar formulations are!
52)
53)
Equating The Lunar And Solar Formulations Yield Our 1 Second Base Unit
Let us equate equation 32 with equation 52:
32)
52)
This gives:
ℏ
⊙
= 9.13E 38J ⋅ s
r
3
=
(9.13E 38)
2
(6.67408E − 11)(5.972E24kg)
3
⋅
1
400.5986
= 1.4638E11m
1.4638E11m
1.496E11m
100 = 97.85 %
−
ℏ
2
2m
[
1
r
2
∂
∂r
(
r
2
∂
∂r
)
+
1
r
2
sinθ
∂
∂θ
(
sinθ
∂
∂θ
)
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
]
ψ + V(r)ψ = E ψ
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
r
n
=
ℏ
2
⊙
GM
3
e
⋅
R
m
R
⊙
ℏ
⊙
= 9.13E 38J ⋅ s
h
⊙
= 2π ℏ
⊙
= 5.7365E 39J ⋅ s
K E
e
= n
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
K E
e
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2ℏ
2
⊙
3
R
⊙
R
m
⋅
G
2
M
2
e
M
3
m
2ℏ
2
⊙
=
R
⊙
R
m
⋅
G
2
M
4
e
M
⊙
2L
2
p
of 21 44
55)
We remember that
This gives
56)
We have
57)
This equates the orbital velocity of the Moon with the centripetal acceleration of the Earth in
terms of one second by way of the mass of the Earth, the mass of the Sun, the mass of the Moon,
and the orbital number of the Earth. Let us compute
58)
Let us see how well equation 57 works. at aphelion is 966 m/s and .
. We have
That is an accuracy of
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
ℏ
⊙
= (hC )K E
p
hC = 1secon d
C =
1
3
⋅
1
α
2
c
1
3
⋅
2π r
p
G m
3
p
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon ds
r
e
v
m
M
e
=
1
6α
2
⋅
r
p
m
p
h 4π
Gc
⋅
1
2
M
e
v
2
e
⋅
M
2
e
M
⊙
M
3
m
3
2v
m
=
v
2
e
r
e
(1secon d )
M
2
e
M
⊙
M
3
m
3
M
2
e
M
⊙
M
3
m
3
=
(5.972E 24kg)
2
(1.9891E 30kg)
(7.34763E 22kg)
3
(1.732)
= 321,331.459 ≈ 321,331
v
m
v
e
= 29,800m /s
r
e
= 1 AU = 1.496E11m
2(966m /s) =
(29,800m /s)
2
1.496E11m
(1sec)(321,331.459)
1,907m /s = 1,932m /s
of 22 44
Equation 57 can be written:
59)
From equation 1:
We have
60)
Since , the diameter of the Earth orbit, we have
61)
And we see the Earth/Moon/Sun system determines the radius and mass of the proton and vice
versa. We basically have
62)
Where is the Earth orbital number. We have
1907
1932
= 98.7 %
1secon d = 2r
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
1
6α
2
m
p
h 4π r
2
p
Gc
= 1.004996352secon ds
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 2r
e
v
m
v
2
e
⋅
M
3
m
3
M
2
e
M
⊙
2r
e
= d
e
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= d
e
v
m
v
2
e
⋅
M
3
m
3
M
2
e
M
⊙
1secon d = 2v
m
r
e
v
2
e
M
3
m
3
M
2
e
M
⊙
1secon d =
K E
moon
K E
earth
Ear th Da y
1secon d =
M
m
v
2
m
M
e
v
2
m
Ear th Da y
Ear th Da y =
2r
e
v
m
M
m
M
⊙
n
n = 3
of 23 44
= 7.83436E4seconds
EarthDay=(24)(60)(60)=86400 seconds,
accuracy
This last equation, equation 62, we can use to find the rotation period, or the length of the day,
of an earth-like planet in the habitable zone of any star system, so it would be very useful.
We want to turn our attention to equation 59 and write it
63)
We see equating solar and lunar formulations for energy yield the base unit of one second.
Equating the lunar and solar solutions for orbitals instead of the for energies, which we just did,
we have
Yields
64)
Where . The accuracy of this is
Thus the energy equations gave the equation 55:
Ear th Da y =
2(1.496E11m)
966m /s
7.34763E 22kg
1.9891E 30kg
1.732
7.834E4s
86,400s
100 = 90.675 %
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
2ℏ
2
⊙
GM
3
m
⋅
R
⊙
R
m
⋅
1
n
=
L
2
p
GM
3
e
⋅
R
m
R
⊙
ℏ
2
⊙
M
3
e
M
3
m
=
3
2
L
2
p
R
2
m
R
2
s
3/2 = cos(π /6)
(2.8314E 33)
2
(5.972E 24)
3
(7.34763E 22)
3
= (0.866)(9.13E 38)
2
(1737400)
2
(6.96E8)
2
4.29059E 72 = 4.5005E 72
4.29059E 72
4.5005E 72
100 = 95.3 %
of 24 44
55) !
And equating the orbital equations gives
65)
These last two yield
66)
The accuracy is
Is 98% accuracy. The important thing that comes out of this is our base unit of a second because
we see it is also a function lunar, solar, and earth masses.
63)
Characteristic Time of the Electron
We have a characteristic time for the proton of one second, we want the characteristic time of an
electron. In modern physics the electron is a point charge and no spatial extent. But you can
associate a size with it that characterizes electron interaction in problems on the atomic scale.
It’s classical radius is
67)
This is several times larger than the proton’s radius. We can write it
68)
Where is the reduced Compton wavelength of the electron. It is
69)
L
p
=
M
2
e
M
⊙
M
3
m
3
⋅ ℏ
⊙
L
2
p
=
2 3
3
⋅
M
3
e
M
3
m
R
2
⊙
R
2
m
⋅ ℏ
2
⊙
2
R
⊙
R
m
M
e
M
⊙
= 1
(400.5986)
5.972E 24kg
1.9891E 30kg
= 1
0.98 = 1
1secon d = d
e
v
m
v
2
e
M
3
m
3
M
2
e
M
⊙
r
e
=
1
4
πϵ
0
⋅
e
2
m
e
c
2
= 2.81794E − 15m
r
e
= λ
e
α = a
0
α
2
λ
e
λ
e
=
ℏ
m
e
c
of 25 44
is the Bohr radius is
70)
We remember our spacetime operators yield the radius of a proton
20)
14).
21).
This is close to the CODATA value for the proton radius around 2018 (0.842E-15m). It may be
the radius of a proton is
We find it may be that the radius of a proton is actually
22)
is approximated by 2/3 which are ratio between two successive fibonacci numbers, where the
fibonacci ratios in its sequence converge on the at infinity. The version for the radius of a
proton using 2/3 could be a characteristic radius of the proton in solving certain kinds of
problems like we have for the electron, but the version using could be the radius-type as given
for the proton when we speak of its radius without specifying what type of radius. The radius of
a proton is not actually precisely defined because it is more a cloud of subatomic particles. Thus
we could have
Radius of Electron
23)
Radius of proton
24).
The Bohr radius would be related to our equations 26 and 36
a
0
a
0
=
4πϵ
0
ℏ
2
e
2
m
e
=
1
α
ℏ
m
e
c
2
3
⋅
π r
p
α
4
G m
3
p
1
3
⋅
h
c
= 1secon d
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
r
p
=
2
3
⋅
h
cm
p
r
p
= ϕ ⋅
h
cm
p
= 0.816632E − 15m
ϕ
ϕ
ϕ
r
e
=
α
2π
⋅
h
m
e
c
r
p
= ϕ ⋅
h
m
p
c
= 0.816632E − 15m
of 26 44
70)
26) Where
36)
The mass of an electron is 9.1093837E-31 kg. Using our space time operator that gives the
characteristic time of the proton as 1 second (equation 14), we can find the characteristic time of
the electron:
14)
We have
We recall
So the characteristic time of an electron is
I find to figure out what this means we have to formulate an intermediate mass and
intermediate velocity somewhere between the higher and lower masses of stars and protons and
the higher and lower velocities in the Universe. We move on to that now…
Intermediate Mass and Velocity
Here I hypothesize an intermediate mass and intermediate velocity (1/k) characteristic of the
Universe. As well I propose an energy characteristic of the Universe. It is suggested that the
characteristic energy of the Universe is related to the kinetic energy of our Moon traveling at the
intermediate velocity.
71)
6 is for sixfold symmetry and square root 3 is the square root of the Earth’s orbital number. 6 is
the atomic number of carbon, upon which the periodic table is based. Because it is 6 protons,
a
0
=
4πϵ
0
ℏ
2
e
2
m
e
=
1
α
ℏ
m
e
c
λ
moon
=
ℏ
2
⊙
GM
3
m
λ
moon
c
=
ℏ
2
⊙
GM
3
m
⋅
1
c
= 1.0secon ds
(
1
6α
2
4πh
Gc
)
⋅
r
p
m
p
= 1secon d
r
e
m
e
=
2.81794E − 15m
9.1093837E − 31kg
= 3.093447E15m /kg
(
1
6α
2
4πh
Gc
)
= 2.018E − 12
kg ⋅ s
m
≈ 2E − 12
kg ⋅ s
m
(3E15)(2E − 12) = 4000secon d s
h
G
⋅
c
3
m
p
=
6
3
M
m
(
1
k
)
2
of 27 44
and 6 neutrons, it is the most stable reference element. The periodic table is periodic over 18
groups is 3 times 6. We have
72)
Our equation has an accuracy of
We have said earlier that the kinetic energy of the Moon is
is a basic energy characteristic of the Universe. is the mass of the Earth’s moon, is a
constant characteristic of the Universe in terms of inverted velocity.
The constant k is given by the integer 6 for sixfold symmetry, and the orbital velocity of the
Earth. It is
73)
But the constant is can also be derived by hypothesizing an intermediate mass characteristic of
the Universe that is the geometric mean between the mass of a proton and the Chandrasekhar
limit for a white dwarf star to not collapse into a blackhole:
74)
The Chandrasekhar limit is given by
75)
76)
Where we have approximated 0.77~3/4 in equation 75.
77)
h
G
⋅
c
3
m
p
= 1.599298E 29J
2 3M
m
(
1
k
)
2
= 2 3(7.347673E 22kg)(788.4626m /s)
2
= 1.58235E 29J
1.58235E 29J
1.599298E 29J
= 98.94 %
K E
m
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
M
m
k
k v
e
= 6
k
m
i
= Mm
p
M ≤ 0.77
c
3
ℏ
3
G
3
N
m
4
p
= 1.41 ⊙
m
i
=
3
2
c
3
ℏ
3
G
3
m
2
p
1/2
k =
4
3
m
p
8π
3
G
c
3
h
(1 + α) ⋅ N
A
𝔼 =
1
773.5
s
m
of 28 44
78)
79)
80)
is any element and is the number of protons in times Avogadro’s number per that
number of grams. would be carbon is 6 grams per 6 protons.
We can hone this by reintroducing 0.77 for 3/4
Thus precisely:
We have
We have honing our
81)
Thus since we said with our estimate
we have a honed value of
82)
83)
To derive this we introduce Giordano’s relationship:
N
A
= 6E 23
proton s
gra m
ℍ = 1
gra m
proton
N
A
⋅ 𝔼 = 6 E 23
𝔼
N
A
𝔼
ℂ
3
2
= 0.8660
0.77 = 0.8775
0.8775
0.8660
= 1.01328
m
i
m
i
= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg
k =
1
m
2
i
h
(1 + α)
G
⋅ N
A
𝔼 =
1
773.5
⋅
s
m
k =
1
(68.897kg)
2
(6.62607E − 34) ⋅
(1.007299)
6.67408E − 11
⋅ 6.02E 23 = 0.001268291s /m
1
k
= 788.4626
m
s
h(1 + α) ⋅ 10
23
= G
of 29 44
The number on the left is the number on the right, but with different units. I found I could
eliminate the and gain six-fold symmetry by creating an equation of state for the periodic
table of the elements using Avogadro’s Number ( ). We say that
And,
Which is basically true, hydrogen is one proton and the mass of the electron is nominal. For
every 6E23 protons of hydrogen, there is one gram. Then we always have
We can say for any element
where is the number of protons in the element, so for carbon
Thus we have
84).
And we have introduced our 6 of six-fold symmetry. We have that
85)
If we substitute for the mass of the Moon our intermediary mass , we get the life span of the
universe in the Friedmann model of the Universe, which is 1E14 years:
10
23
6.02E 23
N
A
= 6E 23
proton s
gra m
ℍ = 1
gra m
proton
N
A
⋅ 𝔼 = 6 E 23
𝔼
N
A
=
Z ⋅ 6E 23protons
Z ⋅ gr a m s
𝔼 =
Z ⋅ gr a m s
Z ⋅ proton s
Z
ℂ
ℂ =
6gra m s
6proton s
N
A
=
6(6E 23pr oton s)
6gra m s
N
A
ℂ = 6 E 23
h
(1 + α)
G
⋅ N
A
𝔼 = 6.00kg
2
⋅
s
m
h
G
⋅
c
3
m
p
2 3M
m
(
1
k
)
2
(1secon d ) = (1secon d )
m
i
of 30 44
86)
Where,
87)
Characteristic Time Of Electron Gives The Moon’s Diameter
Since we have found the solar system is quantized according the base unit of a second the
Moon’s mass, we are hoping the characteristic time of the electron gives the size of the Moon.
We have
The characteristic time of the electron is
And the intermediate velocity is
83)
Thus we have
The diameter of the Moon is 3,474 km. So we do get the diameter of the moon with an accuracy
of
h
G
⋅
c
3
m
p
m
i
(
1
k
)
2
(1secon d ) = Li feSpanUniverse
1
6α
2
⋅
r
p
m
p
h 4π
Gc
= 1secon d
(3E15)(2E − 12) = 4000secon d s
1
k
= 788.4626
m
s
(788.4626
m
s
)(4000s) = 3.254E6m = 3,154k m
3,154k m
3,474k m
100 = 91 %
of 31 44
Biological Life Part of a Universal Natural Process In The Universe
By
Ian Beardsley
Copyright © 2024
of 32 44
Contents
Introduction……………………………………………………………………………………….33
A Theory For Biological Hydrocarbons………………………………………………….34
The Principle of Least Action………………………………………………………………..36
Our Theory For Hydrocarbons………………………………………………………………40
The Data For Verifying The Equations……………………………………………………43
of 33 44
Introduction I have a theory for the star systems, the planets and their suns, wherein such
systems are solved with the Schrödinger wave equation that is used to solve atomic systems. The
result is that star systems and atomic systems, systems on the macro scales and micro scales, are
governed by the same underlying principles. It came to pass that this theory indicated an
overlap with biological systems indicating that biological life could be part of the same idea, and
that is what I hope to pull out of that paper and develop here.
The theory for the planets, which also predicted the radius of a proton, which is one of the
fundamental particles out of which matter is made, suggested that star systems which support
life are a part of a Universal natural process. One of the conditions for star systems that support
life made use of the interesting fact that has mystified science for some time now, that the Moon
is the right size and distance from the Earth that it nearly perfectly eclipses the Sun. The reason
a moon would be necessary for life to be optimally successful on a habitable planet is that we
know our moon that orbits our Earth, makes life here successful because it holds the Earth at its
inclination to its orbit around the Sun allowing for the seasons, thus preventing extreme hot and
extreme cold.
Interestingly, the solution for the planets of the Schrödinger wave equation, was quantized in
terms of the Earth’s moon, and a base unit of one second. The Moon seems to be some sort of a
natural yardstick. The strange thing is that the basis unit of time is one second and the second
was not developed with the planets and atoms in mind, so it is strange that it lands upon the
function of a natural constant in our theory. In the theory (An Abstract Theory of Reality,
Beardsley 2024) I talk a little about why that might be. The second came to us from the Ancient
Sumerians, who invented civilization with settling down from following the herds and hunting
with stone spearpoint to invent agriculture, metallurgy, writing, and mathematics. They had a
base 60 counting system, and found it convenient then to divide the Earth day (rotation period)
into 24 hours, which in the end became adopted by the world. Their base 60 counting resulted in
the Babylonians dividing the hour into 60 minutes, and that in turn into 60 seconds, and that is
how we have the duration of a second we have today. They chose base 60 (sexagesimal) because
60 is evenly divisible by so much from which they gave us the 12 hour day and 12 hour night or
24 hour day from sunrise to sunrise, sunset to sunset. 12 times 5 is 60, also 60 times 6 is 360,
they gave us the 360 degree circle as well.
In this paper I show that the same unit of a second that describes planetary systems and atomic
system in common describes hydrocarbons, the skeletons of biological life chemistry. I further
show that it predicts the atomic radii of the hydrogen and carbon atoms from which such
skeletons are made. Hydrogen and carbon are the most abundant elements in life chemistry,
carbon is the core element upon which life is made, and hydrogen is the simplest element,
element one in the periodic table of the elements, consisting of 1 proton, and is the most
abundant element in the universe by far, and plays the dominant role in all of chemistry.
of 34 44
A Theory For Biological Hydrocarbons I have found that the basis unit of one second is
not just a Natural constant for physical systems like the atom and the planets around the Sun
(See my paper An Abstract Theory of Reality, Beardsley 2024) but for the basis of biological
life, that it is in the sixfold Nature of the chemical skeletons from which life is built, the
hydrocarbons. I found
1).
2).
Where is the fine structure constant, is the radius of a proton, is the mass of a
proton, is Planck’s constant, is the constant of gravitation, is the speed of light, is the
kinetic energy of the Moon, is the kinetic energy of the Earth, and is the
rotation period of the Earth is one day.
The first can be written:
3).
4).
From which instead of saying the left sides of these equations are seconds, we say they are
proton-seconds by not letting the units of cancel with the bodies of these equations on the
left, which are in units of mass, but rather divide into them, giving us a number of protons. I say
this is the biological because these are the hydrocarbons the backbones of biological chemistry.
We see they display sixfold symmetry. I can generate integer numbers of protons from the time
values from these equations for all of the elements with a computer program. Some results are:
1secon d =
1
6α
2
⋅
r
p
m
p
4πh
Gc
1secon d =
K E
m
K E
e
(Ear th Da y)
α = 1/137
r
p
m
p
h
G
c
K E
m
K E
e
Ear th Da y
1
α
2
m
p
h 4π r
2
p
Gc
= 6proton ⋅ secon ds = carbon(C )
1
α
2
m
p
h 4π r
2
p
Gc
= 1proton ⋅ 6secon d s = hydrogen(H )
m
p
of 35 44
A very interesting thing here is looking at the values generated by the program, the smallest
integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds
produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest of
the elements heavier than carbon are formed by fractional seconds. These are the hydrocarbons
the backbones of biological chemistry. And carbon is the core element of life. We see the
duration of the base unit of measuring time, 1 second, given to us by the ancients (the base 60,
sexagesimal, system of counting of the Sumerians who invented math and writing and started
civilization), is perfect for the mathematical formulation of life chemistry. Here is the code for
the program, it finds integer solutions for time values, incremented by the program at the
discretion of the user:
#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
float value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,
c=299792458,protons[100],r=0.833E-15;
do
{
printf("By what value would you like to increment?: ");
scanf("%f", &increment);
printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");
scanf("%i", &n);
}
while (n>=101);
{
for (int i=0; i<n;i++)
{
protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));
int intpart=(int)protons[i];
float decpart=protons[i]-intpart;
t=t+increment;
if (decpart<0.25)
{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);
}}}}
We have that
Since this 6 seconds is also proton-seconds we have
5). is carbon (C)
6). is hydrogen (H)
1
α
2
⋅
r
p
m
p
4πh
G c
= 18769
0.833E − 15
1.67262E − 27
4π (6.62607E − 34)
(6.67408E − 11)(299,792458)
= 6.029978s ≈ 6s
1
6pr oton s
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 1secon d
1
1pr oton
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 6secon d s
of 36 44
Because this is the biological part of a theory of everything in physics (An Abstract Theory of Reality,
Beardsley 2024) we want to talk a little about this physics for biologists.
The Principle of Least Action Action is the Lagrangian, , is the difference in kinetic and
potential energy:!
!
The path, , of a mass from the initial time to the final time is the integral of action over time:!
!
The trajectory it takes is the Path of Least Action, which is not a derived or proven law of
Nature, but a principle of Nature that underlies all of Nature and is at the basis of physics, it is
Newton’s Second Law, , for every action there is an equal and opposite reaction,
where the force is the acceleration of the mass. The principle of least action was first put
forward by Louis Maupertuis 1698-1759 a French mathematician and philosopher. He said:!
The laws of movement and of rest deduced from this principle being precisely the same as
those observed in nature, we can admire the application of it to all phenomena. The movement
of animals, the vegetative growth of plants… are only its consequences; and the spectacle of
the universe becomes so much the grander, so much more beautiful, the worthier of its Author,
when one know that a small number of laws, most wisely established, suffice for all movements.
We can explain this by taking a small change in the path :!
!
The Lagrangian then changes with a small change in path:!
!
Integrate the first part by parts, where integration by parts comes from the product rule for
differentiation.!
!
Integrating both sides and rearranging we have integration parts is:!
!
!
L
L =
1
2
m
(
d x
dt
)
2
− U(x)
S
S =
∫
t
f
t
i
L dt
F = m a
ϵ
x(t) → x(t) + ϵ(t)
d L = m
d x
dt
dϵ
dt
−
dU
d x
ϵ
(u(x)v(x))′ = v(x)u′ (x) + u(x)v′ (x)
∫
u(x)dv = u(x)v(x) −
∫
v(x)du
d
dt
(
m
d x
dt
ϵ
)
= m
d
2
x
dt
2
ϵ + m
d x
dt
dϵ
dt
of 37 44
We have!
!
We take the integral of this for the action!
!
And we set this equal to zero to get the minimum, the shortest path. In the second term epsilon
vanishes at t_i and t_f because it is very small and even smaller because we take the
derivative of it, so the second term goes to zero. We have!
!
For this to be zero we must have!
!
And this is Newtons second Law, . Let us see how we can use the principle of least
action to predict a law of Nature. Here we consider the angle of incidence equals the angle of
reflection.!
!
We want the path from A to B off
some point P.!
The path of least action is the
shortest path, is to set the derivative
of the path equal to zero. The path
is!
!
!
The derivative of its path is!
d L = − m
d
2
x
dt
2
ϵ −
dU
d x
ϵ +
d
dt
(
m
d x
dt
ϵ
)
dS = −
∫
t
f
t
i
(
m
d
2
x
dt
2
−
dU
d x
)
ϵ +
d
dt
(
m
d x
dt
ϵ
)
dt
dS = −
∫
t
f
t
i
(
m
d
2
x
dt
2
−
dU
d x
)
ϵ dt = 0
(
m
d
2
x
dt
2
−
dU
d x
)
= 0
F = m a
d
1
+ d
2
= f (x) = a
2
+ x
2
+ b
2
+ (c − x)
2
of 38 44
and we set it equal to zero:!
, which is , which means !
Thus the path of least action is when the angle of incidence equals the angle of reflection.!
!
Let us find the law of
refraction called Snell’s Law.
By the principle of least
action the path from A to B
refracted by the line
between two mediums, say
air and water, is the path of
least time, which is the
shortest path. Let us say t is
time, c is the speed of light
in air, and v is the speed of
light in water. We have!
!
!
By the principle of least action dt/dx=0, we have!
!
Which is!
!
Which is Snell’s law for refraction.!
f′ (x) =
x
a
2
+ x
2
+
(c − x))(−1)
b
2
+ (c − x)
2
=
x
d
1
−
c − x
d
2
x
d
1
=
c − x
d
2
= 0
cos(α) = cosβ
α = β
t(x) =
x
2
+ a
2
c
+
b
2
+ (s − x)
2
v
dt
d x
=
2x
2 c x
2
+ a
2
+
2(s − x)(−1)
2v b
2
+ (s − x)
2
1
c
⋅
x
d
1
=
1
v
⋅
s − x
d
2
sin(α)
c
=
sin(β )
v
of 39 44
We now only need to understand time. Time is the direction through which events are
separated as space moves through it at the speed of light c.!
!
We consider a particles, say a proton, as pictured
at left in space. As it moves through space (x) it
actually travels through times (t) as well so it
really moves the distance d. We cannot measure
the distance t with a yardstick because our mind
cannot conceive of this direction, so we measure
it with clocks, which are periodic, that have a
frequency, cycles per second. Like an oscillator,
or the Earth’s rotation, which is one rotation
(cycle) per 24 hours, is its frequency.!
Thus as the particle move through x, it moves through a distance t, of time. But how do we
consider time a distance? We have either the distance d is!
Or!
!
The latter because distance is velocity times time and is the distance the particle has moved
through time. We set the two equal and get!
And it says!
!
Thus we have!
Which can be written!
Since , where !
Since we don’t perceive ourselves as moving through the time direction we find the faster we
go, the more time slows down in our inertial frame, and the more distance dilates. We take the
conjugate of this last equation and we have the factor by which time and space dilate in
Einstein’s theory of relativity:!
d = t
2
+ x
2
d = c
2
t
2
+ x
2
ct
t
2
+ x
2
= c
2
t
2
+ x
2
t = ct
d = c
2
t
2
+ x
2
d = 1 +
x
2
c
2
t
2
= 1 +
v
2
c
2
x /t = v
v = velocit y
of 40 44
!
!
Our Theory For Hydrocarbons With all that has been said we are equipped to proceed. We
want to consider the radius of a hydrogen atom and the radius of a carbon atom. The radius of
a carbon atom given in your periodic table of the elements is often 70 to 76 picometers. The
covalent radius of hydrogen is given as 31 picometers. The atomic radius of hydrogen is 53
picometers and the atomic radius of carbon is 67 picometers. We want to consider the atomic
radii of both, because the covalent radius, determined by x-ray diffraction for diatomic
hydrogen, is the size of two hydrogen atoms joined H2 divided by two, where it is measured
that way, joined, in the laboratory. Carbon is C2 divided by 2. We are interested in the single
carbon and hydrogen atoms, because we want to know what our theory for their six-fold
symmetry with one another in their representations in proton-seconds says about the way they
combine as the skeletons of life chemistry. We start with the Planck constant, which is like
flux, a mass (perhaps a number of particles) per second over an area. That is it is kilograms per
second over square area:!
!
We have equations 5 and 6:!
5). is carbon (C)
6). is hydrogen (H)
We can write these
=
7).
=
8). !
t′ =
t
1 −
v
2
c
2
L′ = L 1 −
v
2
c
2
h,
h = 6.62607E − 34J ⋅ s = 6.62607E − 37
kg
s
⋅ m
2
1
6pr oton s
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 1secon d
1
1pr oton
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 6secon d s
(
6.62607E − 37
kg
s
⋅ m
2
)
6secon d s
m
p
(
6.62607E − 37
kg
s
⋅ m
2
)
6seconds
1.67262E − 27kg
= 2.37689E − 6m
2
(
6.62607E − 37
kg
s
⋅ m
2
)
1second
6m
p
(
6.62607E − 37
kg
s
⋅ m
2
)
1second
6(1.67262E − 27kg)
= 6.602486E − 8m
2
of 41 44
We have from 7 and 8…!
9). !
We find the ratio between the surface areas of the hydrogen and carbon atoms:!
10). !
11). !
!
It is the golden mean. These atomic radii are the radii between the nucleus of the atoms and
their valence shell, which is what we want because the valence shell is the outermost electrons
responsible for the way the hydrogen and the carbon combine to make hydrocarbons. We will
write this!
12). !
I am guessing the reason we have the golden mean here is that it is the number used for
closest packing. But what we really want to do is look at the concept of action, for hydrogen
given by six seconds and carbon given by 1 second. We take equations 5 and 6:!
5). is carbon (C)
6). is hydrogen (H)
And, we write
13).
Where is the radius of the atom, and t its time values given here by equations 5 and 6. We have for
hydrogen
14).
=
This is actually very close to the radius of a hydrogen which can vary around this depending on how you
are looking at it, which we said is given by 5.3E-11m. For hydrogen we have:
h
m
p
⋅
(6seconds)
1
h
m
p
⋅
(1second)
6
=
2.37689E − 6m
2
6.602486E − 8m
2
= 35.9999 ≈ 40
H
surface
= 4π (r
2
H
) = 3.52989E − 20m
2
C
surface
= 4π (r
2
C
) = 5.64104E − 20m
2
4π(53pm)
2
4π(67pm)
2
= 0.62575 ≈ 0.618... = ϕ
r
2
H
r
2
C
= ϕ
HC
1
6pr oton s
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 1secon d
1
1pr oton
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 6secon d s
m
p
G c
h
∫
t
0
dt = r
A
r
A
m
p
G c
h
∫
t
0
dt = (1.67262E − 27kg)
(6.67408E − 11)(299,792,458)
6.62607E − 34
∫
6sec
0
dt
(9.2E − 12m /s)(6seco n d s) = 5.5E − 11m
of 42 44
15).
=
And this is actually very close to the radius of a carbon atom which is 6.7E-11m. The thing is if we consider
the bond length of the simplest hydrocarbon CH4, methane, which can be thought of as
16).
our, equations give
17).
which are the same thing. Thus we have the basis for a theory of everything in that it includes the macro
scale, the Earth/Moon/Sun System, because equation 2 gives 1 second:
2).
and it includes the radius of the proton which is the radius and mass of a proton and gives 1
second
1).
and we have the hydrocarbons the skeleton of the chemistry of life give one second
5). is carbon (C)
6). is hydrogen (H)
and because they predict the radii of the Carbon and hydrogen atoms at the core of life
14).
=
15).
=
m
p
G c
h
∫
t
0
dt = (1.67262E − 27kg)
(6.67408E − 11)(299,792,458)
6.62607E − 34
∫
6sec+1sec
0
dt
(9.2E − 12m /s)(7secon d s) = 6.44E − 11m
r
H
+ r
C
= 5.3E − 11m + 6.7E − 11m = 1.2E − 11m
r
H
+ r
C
= 5.5E − 11m + 6.44E − 11m = 1.2E − 11m
1secon d =
K E
m
K E
e
(Ear th Da y)
1secon d =
1
6α
2
⋅
r
p
m
p
4πh
Gc
1
6pr oton s
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 1secon d
1
1pr oton
⋅
1
α
2
⋅
r
p
m
p
4πh
G c
= 6secon d s
m
p
G c
h
∫
t
0
dt = (1.67262E − 27kg)
(6.67408E − 11)(299,792,458)
6.62607E − 34
∫
6sec
0
dt
(9.2E − 12m /s)(6seco n d s) = 5.5E − 11m
m
p
G c
h
∫
t
0
dt = (1.67262E − 27kg)
(6.67408E − 11)(299,792,458)
6.62607E − 34
∫
6sec+1sec
0
dt
(9.2E − 12m /s)(7secon d s) = 6.44E − 11m
of 43 44
The Data For Verifying The Equations
(Proton Mass)
(Planck Constant)
(Proton Radius)
(Gravitational Constant)
(light speed)
(Fine Structure Constant)
Earth day=(24)(60)(60)=86,400 seconds. Using the Moon’s orbital velocity at aphelion, and
Earth’s orbital velocity at perihelion we have:
m
P
: 1.67262 × 10
−27
kg
h : 6.62607 × 10
−34
J ⋅ s
r
p
: 0.833 × 10
−15
m
G: 6.67408 × 10
−11
N
m
2
kg
2
c : 299,792,458m /s
α : 1/137
q
p
= q
e
= 1.6022E − 19coulom bs
k
e
= 8.988E 9
Nm
2
C
2
K E
moon
=
1
2
(7.347673E 22kg)(966m /s)
2
= 3.428E 28J
K E
earth
=
1
2
(5.972E 24kg)(30,290m /s)
2
= 2.7396E 33J
of 44 44
The Author
